Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k + 6}{-3k^2 + 15k + 150} \div \dfrac{k - 8}{k^2 - 18k + 80} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k + 6}{-3k^2 + 15k + 150} \times \dfrac{k^2 - 18k + 80}{k - 8} $ First factor out any common factors. $q = \dfrac{k + 6}{-3(k^2 - 5k - 50)} \times \dfrac{k^2 - 18k + 80}{k - 8} $ Then factor the quadratic expressions. $q = \dfrac {k + 6} {-3(k - 10)(k + 5)} \times \dfrac {(k - 10)(k - 8)} {k - 8} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(k + 6) \times (k - 10)(k - 8) } { -3(k - 10)(k + 5) \times (k - 8)} $ $q = \dfrac {(k - 10)(k - 8)(k + 6)} {-3(k - 10)(k + 5)(k - 8)} $ Notice that $(k - 10)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {\cancel{(k - 10)}(k - 8)(k + 6)} {-3\cancel{(k - 10)}(k + 5)(k - 8)} $ We are dividing by $k - 10$ , so $k - 10 \neq 0$ Therefore, $k \neq 10$ $q = \dfrac {\cancel{(k - 10)}\cancel{(k - 8)}(k + 6)} {-3\cancel{(k - 10)}(k + 5)\cancel{(k - 8)}} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $q = \dfrac {k + 6} {-3(k + 5)} $ $ q = \dfrac{-(k + 6)}{3(k + 5)}; k \neq 10; k \neq 8 $